Answer
(a)
Vertex: $V(\frac{3}{2},-\frac{3}{2})$
Focus: $F(\frac{3}{2},-\frac{3}{4})$
Directrix: $y=-\frac{9}{4}$
(b)
Work Step by Step
Equation of a parabola with vertical axis and vertex at $(h,k)$:
$(x-h)^2=4p(y-k)$
$x^2=3(x+y)$
$x^2-3x=3y$
$x^2-3x+\frac{9}{4}=3y+\frac{9}{4}$
$(x-\frac{3}{2})^2=3[y-(-\frac{3}{2})]$
$h=\frac{3}{2}$
$k=-\frac{3}{2}$
Vertex: $V(h,k)=V(\frac{3}{2},-\frac{3}{2})$
$4p=3$
$p=\frac{3}{4}$
The given equation can be obtained by shifting
$x^2=3y$
right $\frac{3}{2}$ unit and downward $\frac{3}{2}$ unit. In this equation:
Focus: $F(0,p)=F(0,\frac{3}{4})$
Directrix: $y=-p=-\frac{3}{4}$
Now, shift the focus and the directrix $\frac{3}{2}$ unit to the right and $-\frac{3}{2}$ unit downward:
Focus: $F(\frac{3}{2},-\frac{3}{4})$
Directrix: $y=-\frac{9}{4}$
